How do I solve this equation in the field of complex numbers?: $$|z|^2 - z|z| + z = 0 $$ My solutions are: $$z_1 = 0$$ $$z_2 = -1$$

  • 1
    $\begingroup$ $-1$ is not a solution as you would get $1+1-1\not =0$ $\endgroup$ – Henry Nov 21 '18 at 11:20

$$z=\dfrac{|z|^2}{|z|-1}$$ which is real

If $z>0,|z|=+z$ $$0=z^2-z^2+z\iff z=0$$ which is untenable

If $z\le0,|z|=-z$ $$0=z^2+z^2+z=z(2z+1)$$

  • $\begingroup$ Your first step only works when $|z| \ne 1$, so I think you need to explicitly/separately rule out the case that $|z| = 1$. (This isn't too difficult, but I think it's important, if only because the OP believes that $z = -1$ is a solution.) $\endgroup$ – ruakh Nov 21 '18 at 23:29

Observe that $z=0$ is a solution of the equation

$$|z|^2 - z|z| + z = 0.$$

( $z=-1$ is not a solution !)

Therefore let $z \ne 0$ be a further solution of this equation. We get , since $|z|^2=z \overline{z}:$

$\overline{z}-|z|+1=0$. This gives $\overline{z}=|z|-1 \in \mathbb R$.Hence

$z=|z|-1 \in \mathbb R$.

If $z>0$ , we have $z=z-1$, which is impossible. Hence $z<0$ and then $z=-1/2$.


WLOG $z=r(\cos t+i\sin t)$ where $r>0,t$ are real

$$r(r-r(\cos t+i\sin t)+(\cos t+i\sin t))=0$$

If $r\ne0,$ equating the real & the imaginary parts

$r-r\cos t+\cos t=0=(1-r)\sin t$

Case $\#1:$

If $r=1,1=0$ which is untenable

If $\sin t=0,$

Case $\#2A:\cos t=1,r=0$ which is untenable

Case $\#2A:\cos t=-1,r+r-1=0\iff r=?$


You may also proceed as follows:

  • Rewrite the equation to $$|z|^2 - z|z| + z = 0 \Leftrightarrow \boxed{|z|^2 = z(|z|-1)}$$
  • Therefore, $\color{blue}{z}$ must be $\color{blue}{\mbox{real}}$ and so we have $\color{blue}{|z|^2 = z^2}$.
  • Noting the solution $\boxed{z = 0}$ we get $$|z|^2 = z(|z|-1) \stackrel{z \in \mathbb{R}, z \neq 0}{\Leftrightarrow}z = |z|-1 \Rightarrow \boxed{z = -\frac{1}{2}}$$

Let $r = e^{i\theta}$.

We get $r^2 - re^{i\theta}\cdot r + re^{i\theta} = 0$.

Factorising, we get $r = 0$, giving $z = 0$ as one solution or:

$r - re^{i\theta} + e^{i\theta} = 0$

$e^{i\theta} = \frac{r}{r-1}$

From the last equation, the magnitude of the LHS is equal to one. From the RHS, $e^{i\theta} \in \mathbb{R}$. Hence $e^{i\theta} = \pm 1$.

$\frac{r}{r-1} = 1$ gives no solution, but $\frac{r}{r-1} = -1 \implies r = \frac 12$. Since $e^{i\theta} = -1$ that gives $z = -\frac 12$.

Therefore the two solutions are $z = 0$ and $z = -\frac 12$.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.